IB Chemistry HL

Revision Notes

19.1.3 Electrolysis of Aqueous Solutions

Electrolysis of Aqueous Solutions

  • We have seen previously how simple binary compounds can be electrolysed when molten and the products of electrolysis can be predicted using our knowledge of the ions present
  • At the cathode, positive metals ions (cations) are discharged resulting in metals being deposited
  • The cations are reduced by the electrons coming from the cathode:

Pb2+(l) + 2e  → Pb (l)

  • Meanwhile, at the anode, anions are discharged by oxidation:

2Br (l)  Br2 (g) + 2e

  • However, when aqueous solutions of ionic compounds are electrolysed the products are a little more complicated to predict as there are additional ions present from the water
  • Water can be oxidised or reduced:
    • Oxidation reaction:

2H2O (l) → H2 (g) + 2OH (aq) + 2e

    • Reduction reaction:

2H2O (l) + 2e–  → 4H+ (aq) + O2 (g)

  • At the cathode either the metal ion M+ or water can be reduced
  • At the anode either the anion A- or water can be oxidized
  • Which species is discharged depends on three things:
    • The relative values of Eθ
    • The concentration of the ions present
    • The identity of the electrode


Products of specified electrolytes

  • The electrolysis of water, sodium chloride solution and copper sulfate solutions is as follows:

Table showing the electrolysis products of aqueous solutions









The influence of relative values of Eθ

  • The electrolysis of water is very slow as there are few ions present, so a little acid or base can be added to increase the number of ions present and speed up the electrolysis
  • Whether acid or base is added the products are the same, but the electrode reactions are slightly different
  • Using dilute sulfuric acid as the electrolyte, the cathode reactions could be

2H2O (l) + 2e →   4H+ (aq)  + O2 (g)             Eθ = -0.83V

2H+ (aq) + 2e →  H2 (g)                               Eθ =  0.00 V

  • The Eθ is smaller for the hydrogen ion so it is preferentially reduced and H2 (g) will be discharged
  • At the anode, although sulfate ions are present in the solution, only water can be oxidised
  • This is because the sulfate ion, SO42-, contains sulfur in its maximum oxidation state (+6) so it cannot be further oxidised
  • The oxidation of water produces oxygen gas

2H2O(l)   →  4H+ (aq) +  O2 (g) +  4e          Eθ = +1.23 V

  • If the water is made basic by the addition of dilute sodium hydroxide solution, the cathode reactions could be:

Na+ (aq) + e → Na (s)                                Eθ =  -2.71 V

2H2O (l) + 2e → H2 (g) +  4OH (aq)         Eθ = -0.83 V

  • The Eθ is smaller for water than the sodium ion, so water is preferentially reduced and H2 (g) will be discharged
  • At the anode, either the hydroxide ion or water can be oxidised

4OH(aq)  →  2H2O (l) + O2 (g)  +  4e                    Eθ = +0.40 V

2H2O (l)   →  4H+ (aq) + O2 (g) +  4e                          Eθ = +1.23 V

  • Based on these values the hydroxide ion is preferentially oxidized and O2 (g) will be discharged
  • The overall reaction whether in acid or alkali conditions is:

2H2O (l)  →  2H2 (g)   + O2 (g)

The influence of concentration of the ions

  • The electrolysis of sodium chloride solution provides an illustration of the influence of concentration on the products discharged
  • As before, we would expect hydrogen ion to be preferentially discharged at the cathode before the sodium ion:

2H+ (aq) + 2e → H2 (g)                 Eθ =  0.00 V

  • However at the anode, the relative proximity of the Eθ values allows the possibility of both reactions occurring:

           2Cl–  (aq) → Cl2 (g)  + 2e              Eθ = -1.36 V

    2H2O (l) →  4H+ (aq)  +  O2 (g) +  4e-     Eθ = -1.23 V

  • In fact, when concentration of the sodium chloride increases to more than 25% the Cl becomes preferentially discharged and chlorine gas is the main product of the reaction at the anode
  • The overall reaction equation is

2NaCl (g)  + 2H2O (l) →    2NaOH (aq) +  H2 (g)  +  Cl2 (g)

Influence of the electrodes

  • The products of electrolysis is influenced by the identity of the electrodes
  • Electrodes that take part in the redox processes are know as active electrodes and inert electrodes such as platinum and carbon are called passive electrodes
  • The electrolysis of copper sulfate solution, CuSO4 (aq),  is an example of where active and passive electrodes determine the products

Active electrodes

  • At the cathode, the possible reactions that could take place are:

      Cu2+ (aq) + 2e → Cu (s)                         Eθ =  +0.34 V

  2H2O (l) + 2e → H2 (g) +  4OH (aq)        Eθ = -0.83 V

  • Copper ions are preferentially reduced so copper metal is deposited on the cathode
  • At the anode, water is oxidised, so oxygen gas is produced (the sulfate ion cannot be oxidised)

    2H2O (l) →  4H+ (aq)  +  O2 (g) +  4e     Eθ = -1.23 V

  • The overall equation for the reaction is:

2CuSO4 (aq) + 2H2O (l) → 2Cu (s) + O2 (g) + 2SO42- (aq) 4H+ (aq) 


2CuSO4 (aq) + 2H2O (l) → 2Cu (s) + O2 (g) + 2H2SO4 (aq)

Passive electrodes

  • At the cathode the reaction is the same as with inert electrodes:

Cu2+ (aq) + 2e → Cu (s)                         Eθ =  +0.34 V

  • However, at the anode the copper electrode is oxidised and dissolves to form copper ions

Cu (s)  Cu2+ (aq) + 2e                      Eθ =  -0.34 V

  • This reaction is used to purify copper, needed to produce a very high grade of copper for use in electrical wires
  • The impure copper is made the anode, and the cathode is made of pure copper
  • The impurities fall from the anode fall to the bottom of the cell






The purification of copper by electrolysis


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